The owner of a cannery wants to design a can with a volume of 300 cubic
centimeters. He wants the can to be made of the least amount of material in an
effort to minimize the cost of manufacturing. Find the radius and height of the
can that would satisfy his requirements.
We are asked to minimize the surface area of a closed right circular cylinder,
We need two formulas, one for the volume and one for the surface area of a
closed cylinder. These are
V = pr²h and A = 2prh+2pr²
Since the volume is fixed at 300 cm³, we have
300 = pr²h
Solve this for the height h
h = 300/(pr²)
Now substitute 300/(pr²) for h in the surface area formula
A = 2prh+2pr²
A = 2pr[300/(pr²)] + 2pr²
A = 600/r + 2pr²
Now we take the derivative dA/dr
dA/dr = -600/r² + 4pr
set this = 0
-600/r² + 4pr = 0
Multiply through by LCD = r²
-600 + 4pr³ = 0
4pr³ = 600
r³ = 600/(4p)
r³ = 150/p
` ` ` ``_____
` r = ³Ö150/p = 3.627831679 cm, approximately.
h = 300/(pr²) = 7.255663357 cm, approximately
So the height is twice the radius, making the diameter equal to the height.
If the can were sliced in half vertically the cross section would be a square.
Edwin
